Three line summary

Conditional expectations exist in a natural way for simple functions, by taking extensions they also exist for integrable functions to a Banach space $L^1(\Omega\to E)$.
Using conditional expectations we can define what a martingale is just like in the real case.
The space of continuous $p$-integrable martingales is a Banach space.

Why should I care?

Banach valued martingales form the basis of SPDEs. This is because analogously to Itô integration of real-valued processes. Integrating against a Wiener process valued in a Banach space the same will produce a square-integrable continuous martingale.

1. Conditional expectation

In graduate-level probability courses, given a $\sigma-$algebra $\mathcal{G}$ one shows that by applying Radon-Nikodyn’s theorem, for any real-valued random variable $X\in L^1(\Omega\to R)$ there exists a conditional expectation $\mathbb{E}\U {\mathcal{G}}\zl X\zr $ verifying that

$$\int\U A \mathbb{E}\U {\mathcal{G}}\zl X\zr =\int\U A X,\quad \forall A\in\mathcal{G}.$$

Of course, now that we have created an integral for the space of random variables $L^1(\Omega\to E)$ to some Banach space $E$, we would like to see whether such a conditional expectation also exists for these functions. To start with, if we are given a simple function

$$X=\sum\U {k=1}^n x\U k 1\U {A\U k}, \quad x\U k\in E, A\U k\in\mathcal{G}.$$

It is a simple calculation to show that

$$\mathbb{E}\U {\mathcal{G}}\zl X\zr =\sum\U {k=1}^n x\U k \mathbb{E}\U {\mathcal{G}}\zl 1\U {A\U k}\zr ,$$

verifies the desired formula (we note that, since $1\U {A\U k}$ are real-valued, so $\mathbb{E}\U {\mathcal{G}}\zl 1\U {A\U k}\zr $ are well defined). Furthermore, we have that $\mathbb{E}\U \mathcal{G}$ is a linear, and pointwise continuous operator with

$$\|\mathbb{E}\U {\mathcal{G}}\zl X\zr \|\leq \sum\U {k=1}^n \|x\U k\| \mathbb{E}\U {\mathcal{G}}\zl 1\U {A\U k}\zr =\mathbb{E}\U {\mathcal{G}}\left[ \sum\U {k=1}^n \|x\U k\|1\U {A\U k}\right] =\mathbb{E}\U {\mathcal{G}}\left[ \|X\|\right] .$$

This allows us to show the following

Theorem 1 (Existence and uniqueness of conditional expectation). Let $X\in L^1(\Omega\to E)$ for some Banach space $E$. Then $X$ has a conditional expectation satisfying

$$\|\mathbb{E}\U {\mathcal{G}}\zl X\zr \|\leq\mathbb{E}\U {\mathcal{G}}\left[ \|X\|\right] .$$

Proof. We have already proved the above inequality for simple processes. By the previous post, 1 we can take $X\U n$ converging to $X$ in $L^1(\Omega\to E)$ to obtain that

$$ \|\mathbb{E}\U {\mathcal{G}}\zl X\U n-X\U m\zr \|\leq\mathbb{E}\U {\mathcal{G}}\left[ \|X\U n-X\U m\|\right] \implies \mathbb{E}\zl \|\mathbb{E}\U {\mathcal{G}}\zl X\U n\zr -\mathbb{E}\U {\mathcal{G}}\zl X\U m\zr \|\zr \leq \mathbb{E}\left[ \|X\U n-X\U m\|\right] \to 0. $$

As a result, $\mathbb{E}\U {\mathcal{G}}\zl X\U n\zr $ is a Cauchy sequence in the complete space $L^1(\Omega\to E)$, and so converges to some function $Z$. Now passing to the limit in the defining equation for the conditional expectation shows that $Z=\mathbb{E}\U \mathcal{G}\zl X\zr $. Finally, to prove uniqueness we have that if both $Z\U 1, Z\U 2$ satisfy

$$\int\U A Z\U 1=\int\U A X=\int\U A Z\U 2,\quad \forall A\in\mathcal{G}.$$

Then using the linearity of the Bochner integral we obtain that $w(Z\U 1)=w(Z\U 2)$ for all linear function $w$, so $Z\U 1=Z\U 2$. ◻

In proofs it is often useful to reduce infinite dimensional conditional expectations to finite dimensional ones. This can be done with the following trick

Proposition 1 (Linearity of the conditional expectation). Let $X\in L^1(\Omega\to E)$ for some Banach space $E$. Then it holds that

$$\|\mathbb{E}\U {\mathcal{G}}\zl \ell(X)\zr \|\=\ell\left(\mathbb{E}\U {\mathcal{G}}\left[ \|X\|\right]\right) .$$

Proof. This follows from the definition of conditional expectation and the linearity of the Bochner integral. ◻

2. Martingales

Okay, so we leveraged some inequalities to prove the existence of a conditional expectation. This done, the following definition, which mimics the real case, is quite natural.

Definition 1. Let $\{M(t)\}\U {t\in I}$, be a stochastic process on $(\Omega, \mathcal{F}, \mathbb{P})$ with a filtration $\{\mathcal{F}\U {t}\}\U {t \in I}$. The process $M$ is called an $\mathcal{F}\U {t}$-martingale, if:

$M(t)\in L^1(\Omega\to E)$ for all $t\in I$
$M(t):\mathcal{F}\U {t} \to \mathcal{B}(E)$ for all $t\in I$,
$\mathbb{E}\U {\mathcal{F}\U {s}}\left[ M(t)\right] =M(s)$ for all $s \leq t$.

The concept of submartingale (supermartingale) is defined by replacing the equality in 3. with a $\geq$ (with a $\leq$). In the real case, the absolute value of a martingale is a submartingale. To obtain this result for Banach valued Martingales we need to assume that $E$ is separable so that we can use the following result.

Lemma 1 Let $E$ be a separable metric space, then there exists a countable family of linear functions $\ell \U n \in E^\star $ such that

$$\norm{e}=\sup\U {n\in \mathbb{N}}\ell_n(e)\quad\forall e\in E.$$

Proof. Let $e\U n$ be a countable dense subset of $E$. By Hahn Banach’s theorem we may take $\ell\U n$ such that

$$\norm{\ell\U n}=1;\quad \norm{\ell\U n(e\U n)}=\norm{e \U n} \quad\forall n\in \mathbb{N}.$$

By density, we may take a sequence $e\U {n\U k} \to e$. This gives,

$$\norm{\ell\U {n\U k}(e)}\geq\norm{\ell\U {n\U k}(e\U {n\U k})}-\norm{\ell\U {n\U k}(e-e\U {n\U k})}\geq \norm{e\U {n\U k}}-\norm{e-e\U {n\U k}}.$$

Taking limits we conclude that

$$\sup\U {n\in \mathbb{N}}\ell_n(e)\geq \lim\U {k\to\infty}\norm{\ell\U {n\U k}(e)}=\norm{e}.$$

Since the reverse inequality holds by definition of the norm on $E^\star$ this concludes the proof. ◻

The crucial part of requiring $E$ to be separable is that the norm is a countable supremum of linear functions. We recall that it is the only countable supremum of measurable functions that are measurable. Let us abbreviate $\mathbb{E}\U {\mathcal{F}\U t}$ by $\mathbb{E}\U t$. Then, as in the real case, we have the following.

Lemma 2 Let $M(t)$ be a martingale valued in a separable metric space $E$ , then $\norm{M(t)}$ is a submartingale.

Proof. Let us take $\ell\U n$ as in Lemma 1. Then, since $M$ is a martingale, by the linearity of conditional expectation (Proposition 1) and by Fatou’s lemma for the limsup

$$ \|M(s)\|=\|\mathbb{E}\U {s}\zl M(t)\zr \|= \sup\U {n\in \mathbb{N}} \ell \U n\left(\mathbb{E}\U {s}\zl M(t)\zr \right)=\sup\U {n\in \mathbb{N}}{\mathbb{E}\U {s}\left[ \ell \U n(M(t))\right] } \leq \mathbb{E}\U {s}\left[ \sup\U {n\in \mathbb{N}}\ell \U n(M(t))\right] =\mathbb{E}\U {s}\left[ \|M(t)\|\right]. $$

◻

Let us recall the following result for real-valued martingales

Lemma 3 (Doob’s maximal martingale inequality). Let $\{X\U k\}\U {k=1}^\infty$ be a real-valued sub-martingale. Then it holds that

$$\norm{\max\U {k\in\lb 1,...,n \rb}X\U k}\U {L^p(\Omega)}\leq \frac{p}{p-1}\norm{X\U n}\U {L^p(\Omega)}.$$

As a consequence, if $X\U t,t\in\zl 0,T\zr $ is left (or right) continuous then

$$\norm{\max\U {t\in\zl 0,T\zr }X\U t}\U {L^p(\Omega)}\leq \frac{p}{p-1}\norm{X\U T}\U {L^p(\Omega)}.$$

The idea of the above result is that since $X\U k$ is a submartingale, $X\U k\lesssim X\U {k+1}\lesssim…\lesssim X\U n$. Getting from the continuous to the discrete case is possible by using the continuity of $X$ and approximating it on some finer and finer mesh $t\U 0,…,t\U n$. For all the details see page $5$ of 2.

This said, applying Doob’s maximal martingale inequality together with the Lemma $1$ gives that

Theorem 2 (Doob’s maximal inequality). Let $p>1$ and let $E$ be a separable Banach space. If $M(t)\in L^p (\Omega \to E)$, is a right-continuous $E$-valued $\mathcal{F}\U {t}$-martingale, then

$$ \left(E\left(\sup \U {t \in\zl 0, T\zr }\|M(t)\|^{p}\right)\right)^{\frac{1}{p}} \leq \frac{p}{p-1} \sup \U {t \in\zl 0, T\zr }\left(E\left(\|M(t)\|^{p}\right)\right)^{\frac{1}{p}} =\frac{p}{p-1}\left(E\left(\|M(T)\|^{p}\right)\right)^{\frac{1}{p}}. $$

Proof. This follows by using that $\norm{M(t)}$ is a real valued sub-martingale and Doob’s maximal inequality. ◻

Doob’s inequality is essentially an equality between different function norms we can place on the space of continuous Martingales and will provide a very powerful tool later on. We state this precisely below.

Corollary 1. Let $M$ be a (left or right) continuous martingale to a separable Banach space $E$. Then the following are equivalent

$M\in L^\infty(\zl 0,T\zr \to L^2(\Omega\to E))$.
$M\in \hat{L}^2(\Omega\to L^\infty(\zl 0,T\zr \to E))$.
$\mathbb{E}\zl \norm{M(T)}^2\zr <\infty$.

Where we recall from the previous post that $\hat{L}^p$ symbolizes that $M$ may not be separately valued and only have an integrable norm. That said, the same reasoning shows that the above result also holds for the integrable $L^p$ spaces.

A useful space of Martingales is as follows

Definition 2. Let $M(t)$ be a $E$ valued martingale with index set $I=\zl 0,T\zr $, then we define

$$\mathcal{M}\U T^2(E):=\left\{\text{continuous martingales } M:\mathbb{E}\zl \norm{M(T)}^2\zr <\infty\right\},$$

and give it the norm

$$\norm{M}\U {\mathcal{M}\U T^2(E)}:=\mathbb{E}\zl \norm{M(T)}^2\zr .$$

By Theorem $2$ we have that

$$\mathcal{M}\U T^2(E)\subset L^\infty(\zl 0,T\zr \to L^2(\Omega\to E))\cap \hat{L}^2(\Omega\to L^\infty(\zl 0,T\zr \to E)).$$

And that any of the norms of these spaces is equivalent to the one set on $\mathcal{M}\U T^2(E)$. This is useful in the following result

Proposition 1. Let $E$ be a separable Banach (Hilbert) space, then $\mathcal{M}\U T^2(E)$ is a Banach (Hilbert) space.

Proof. Let us first consider the case when $E$ is a Banach space. By the previous observation and the completeness of the $\hat{L}^p$ spaces proved in the previous post, $\mathcal{M}\U T^2(E)$ is a subspace of a Hilbert space. As a result, it is sufficient to show that it is closed. Let $M\U n$ converge to $M \in \hat{L}^2(\Omega\to L^\infty(\zl 0,T\zr \to E))$. Then, by the equivalence of the norms, we have that $M\U n(t)\to M(t)\in L^1(\Omega\to E)\subset L^2(\Omega\to E)$ so that for all $A\in\mathcal{F}\U s$

$$\int\U A M(s)d\mathbb{P}=\lim\U {n\to\infty}\int\U A M\U n(s)d\mathbb{P}=\lim\U {n\to\infty}\int\U A M\U n(t)d\mathbb{P}=\int\U A M(t)d\mathbb{P}.$$

This shows that $M$ is a martingale. Furthermore, as was seen in Proposition 2 of the previous post 1

$$\lim\U {n\to\infty}{M\U {n\U k}}(\cdot,\omega)=M(\cdot,\omega)\in L^\infty(\zl 0,T\zr \to E)\quad a.e.\quad \omega\in\Omega.$$

Since $M\U {n\U k}(\cdot,\omega)$ are continuous and continuity is preserved by uniform limits this proves that $M$ is continuous almost everywhere. In consequence, $\mathcal{M}\U T^2(E)$ is closed and thus a Banach space. In the case where $E$ is a Hilbert space we can endow $\mathcal{M}\U T^2(E)$ with the inner product

$$\langle{M\U 1, M \U 2\rangle}\U {\mathcal{M}\U T^2(E)}:= \mathbb{E} \left[ \langle{M\U 1(T), M \U 2(T)\rangle}\U E \right ].$$

This concludes the proof. ◻

In future installments, we will prove that a Banach valued Wiener process with trace class covariance belongs to this space and use it to define the stochastic integral that leads to the construction of SPDEs.

Tags: Stochastic calculus